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ira [324]
3 years ago
11

The function g(x) is graphed below. Find g(2).* N 22 need help solving please ​

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
6 0

Answer:

x =2, The function g(x) = g(2) = 1

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Pre calculus law of sines
lakkis [162]

Answer:

AB = 6.5

Step-by-step explanation:

C = 59°

AC = 7

CD = 6

AB = ?

Apply the Law of Cosines, which is AB² = AC² + CD² - 2(AC)(CD)*Cos(C)

AB² = 7² + 6² - 2(7)(6)*Cos(59)

AB² = 85 - 43.26

AB² = 41.74

AB = √41.74

AB = 6.5 (nearest tenth)

6 0
3 years ago
ASAP!!! PLEASE help me with this question!
mihalych1998 [28]

Answer:

C. 16π

Step-by-step explanation:

Well we need to find the area of the top circle which has a radius of 4cm.

π(4)^2

16π in terms of pi

<em>Thus,</em>

<em>the answer is C. 16π.</em>

<em />

<em>Hope this helps :)</em>

4 0
3 years ago
Is 4, 6, 7 a pythagorean triple
madam [21]

Answer:

Yes

Step-by-step explanation:

Hope this helps!

Good Luck!

7 0
3 years ago
Read 2 more answers
Square root of 5 + square root of 3 the whole divided by sqaure root of 5 - square root of 3
xxMikexx [17]

Answer:

The answer is 4 + √15 .

Step-by-step explanation:

You have to get rid of surds in the denorminator by multiplying it with the opposite sign :

\frac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{5}  -  \sqrt{3} }

=  \frac{ \sqrt{5} +  \sqrt{3}  }{ \sqrt{5}  -  \sqrt{3} }  \times  \frac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{5}  +  \sqrt{3} }

=  \frac{ {( \sqrt{5} +  \sqrt{3} ) }^{2} }{( \sqrt{5} -  \sqrt{3} )( \sqrt{5}  +  \sqrt{3})  }

=   \frac{ {( \sqrt{5} )}^{2}  + 2( \sqrt{5} )( \sqrt{3}) +  {( \sqrt{3}) }^{2}  }{ {( \sqrt{5}) }^{2} -  { (\sqrt{3} )}^{2}  }

=  \frac{5 + 2 \sqrt{15} + 3 }{5 - 3}

=  \frac{8 + 2 \sqrt{15} }{2}

= 4 +  \sqrt{15}

3 0
3 years ago
Pls help me with this answer choice question! brainliest, rattings, thanks etc.
galina1969 [7]

Answer:

graph B

Step-by-step explanation:

Given

3y - 5x < - 6

Since the inequality is < then the line must be broken

If ≤ then line would be solid

The only possible graphs are B and D

Choose a test point in the shaded region of each graph and check validity

graph B (2, 0) ← substitute into left side of inequality

0 - 5(2) = 0 - 10 = - 10 < - 6 ← this is valid

graph D (0, 0 ) ← substitute into left side of inequality

0 - 0 = 0 > - 6 ← this is not valid

Hence graph B is the correct graph

3 0
3 years ago
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